The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following transition metal ions has highest magnetic moment?

A
$cu^{ 2+ }$

B
$Ni^{ 2+ }$

C
$Co^{2+}$

D
$Fe^{2+}$

Solution

Magnetic moment depends on number of unpaired electrons,in $Fe^{+2}$ has $4$ unpaired electrons while $Co^{+2}$ , $Ni^{+2}$ , $Cu^{+2}$ has $3,2,1$ unpaired electrons respectively.
Question 2
1 / -0

Select the correct option, among $Sc(III),\ Ti(IV),\ Pd(II)$ and $Cu(II)$ ions.

A
All are paramagnetic

B
All are diamagnetic

C
$Sc(III),\ Ti(IV)$ are paramagnetic and $Pd(III),\ Cu(II)$ are diamagnetic

D
$Sc(III),\ Ti(IV)$ are diamagnetic and $Pd(II),\ Cu(II)$ are paramagnetic

Solution

${ SC }^{ 3+ }\left( { [Ar]3d }^{ 0 } \right)$ ] $\rightarrow$ since no unpaired electrons so diamagnetic

${ Ti }^{ 4+ }\left([Ar] { 3d }^{ 0 } \right)$ ] $\rightarrow$ since no unpaired electrons so diamagnetic

${ Pd }^{ 2+ }\left( [Kr]{ 4d }^{ 8 } \right)$ ] $\rightarrow$ since there is unpaired electrons so paramagnetic

${ Cu }^{ 2+ }\left([Ar] { 3d }^{ 9 } \right)$ ] $\rightarrow$ since there is unpaired electrons so paramagnetic.

So, the correct option is D.
Question 3
1 / -0

$CuSO_4$ is paramagnetic while $ZnSO_4$ is diamagnetic because:

A
$Cu^{2+}$ ion has $3d^9$ configuration while $Zn^{2+}$ ion has $3d^{10}$ configuration.

B
$Cu^{2+}$ ion has $3d^5$ configuration while $Zn^{2+}$ ion has $3d^6$ configuration.

C
$Cu^{2+}$ has half filled orbitals while $Zn^{2+}$ has fully filled orbitals.

D
$CuSO_4$ is blue in colour while $ZnSO_4$ is white.

Solution

$CuSO_4$ has $Cu^{+2}$ ion which has electronic configuration of $[Ar]3d^94s^0$ while in $ZnSO_4$ , $Zn^{+2}$ has $[Ar]3d^{10}4s^0$ configuration.
Question 4
1 / -0

The correct order of a number of unpaired electrons is :

A
$Cu^{2+} > Ni^{2+} > Cr^{3+} > Fe^{3+}$

B
$Ni^{2+} > Cu^{2+} > Fe^{3+} > Cr^{3+}$

C
$Fe^{3+} > Cr^{3+} > Ni^{2+} > Cu^{2+}$

D
$Cr^{3+}> Fe^{3+} > Ni^{2+} > Cu^{2+}$

Solution

The no. of unpaired e $^-$ in the given ions is:
Cr $^{3+}$ - [Ar] 3d $^34s^0$
Fe $^{3+}$ - [Ar] 3d $^54s^0$
Ni $^{2+}$ - [Ar] 3d $^84s^0$
Cu $^{2+}$ - [Ar] 3d $^94s^0$

$Fe^{3+} - 3d^{5}$ No. of unpaired electrons = $5$
$Cr^{3+} - 3d^{3}$ No. of unpaired electrons = $3$
$Ni^{2+} - 3d^{8}$ No. of unpaired electrons = $2$
$Cu^{2+} - 3d^{9}$ No. of unpaired electrons = $1$

The order is: Fe $^{3+}$ > Cr $^{3+}$ > Ni $^{2+}$ >Cu $^{2+}$
Question 5
1 / -0

Which of the following group contains coloured ions?
1. $Cu^+$
2. $Ti^{4+}$
3. $Co^{2+}$
4. $Fe^{2+}$

A
$1,4$

B
$3,4$

C
$2,3$

D
$1,2$

Solution

$Cu^{+}$ and $Ti^{+4}$ have fully filled and empty d orbital respectively. Hence due to absence of unpaired electrons,they will be colourless. But $Fe^{+2}$ and $Co^{+2}$ chave unpaired electrons. Hence they will be coloured.
Question 6
1 / -0

Which of the following transition metal ions is colourless?

A
$V^{2+}$

B
$Cr^{3+}$

C
$Zn^{2+}$

D
$Ti^{3+}$

Solution

The ion will show colour if it has unpaired electrons in its electronic configuration,in $Zn^{+2}$ there are no unpaired electrons due to fully filled d orbital. Hence will be colourless.
Question 7
1 / -0

Potassium dichromate is prepared from:

A
chromate obtained by the fusion of chromite ore with sodium carbonate in free access of air

B
pyrolusite which is fused with potassium hydroxide in the presence of air

C
iron pyrites by the fusion with potassium carbonate in presence of moisture

D
none of these

Solution

The potassium chromate can be prepared by chromate ore $FeCr_2O_4$ .
The various reactions given are:
(i)4 $FeCr_2O_4$ + 4 $Na_2CO_3$ +9 $O_2$ $\rightarrow$ $8NaCrO_4$ + $2Fe_2O_3$ + $4CO_2$ (chromite ore).
(ii) $2Na_2CrO_4$ + $H_2SO_4$ $\rightarrow$ $Na_2Cr_2O_7$ + $Na_2SO_4$ + $H_2O$
(iii) $2Na_2Cr_2O_7$ + $2KCl$ $\rightarrow$ $2K_2Cr_2O_7$ + $2NaCl$
Question 8
1 / -0

Compound that is both paramagnetic and coloured is:

A
$K_2Cr_2O_7$

B
$(NH_4)_2[TiCI_6]$

C
$VOSO_4$

D
$K_3[Cu(CN)_4]$

Solution

In VOSO $_4$ we have no VO $^+$ ion in which oxidation state of V is +3.
So, the electronic configuration for $V^{3+}$ is [Ar] 3d $^24s^0$ because of presence of 2 unpaired e $^-$ s it is paramagnetic and coloured as well.
K $_2Cr_2O_7$ - diamagnetic -orange
(NH $_4)_2$ [TiCl $_6$ ]- diamagnetic - colourless
$K_3[Cu(CN)_4$ ] - diamagnetic- colourless.
Question 9
1 / -0

Transition metals make the most efficient catalysts because of their ability to:

A
adopt multiple oxidation states and to form complexes

B
form coloured ions

C
show paramagnetism due to unpaired electrons

D
form a large number of oxides

Solution

Transition metals have partially filled d- orbitals so they can easily withdraw the electrons from the reagents or give electrons to them depending on the nature of the reaction. They also have a tendency to show large no. of oxidation states and the ability to form complexes which makes them a good catalyst.
Question 10
1 / -0

Most of the transition metals exhibit:
(i) paramagnetic behaviour
(ii) diamagnetic behaviour
(iii) variable oxidation states
(iv) formation of coloured ions

A
(ii), (iii) and (iv)

B
(i), (iii) and (iv)

C
(i), (ii) and (iii)

D
(i), (ii) and (iv)

Solution

Most of the transition metals exhibit:
i) Paramagnetic Behaviour- Because of presence of unpaired e $^-$ s in their orbitals.
iii) Variable oxidation states- Because of vacant d-orbitals
iv) Formation of coloured ions- Because of unpaired e $^-$ s in their orbitals which undergoes d-d transition or charge transfer phenomenon.

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The d- and f- Block Elements Test - 37

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The d- and f- Block Elements Test - 37

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Question 1

cu2+cu^{ 2+ }cu2+

Ni2+Ni^{ 2+ }Ni2+

Co2+Co^{2+}Co2+

Fe2+Fe^{2+}Fe2+

Solution

Question 2

All are paramagnetic

All are diamagnetic

Sc(III), Ti(IV)Sc(III),\ Ti(IV)Sc(III), Ti(IV) are paramagnetic and Pd(III), Cu(II)Pd(III),\ Cu(II)Pd(III), Cu(II) are diamagnetic

Sc(III), Ti(IV)Sc(III),\ Ti(IV)Sc(III), Ti(IV) are diamagnetic and Pd(II), Cu(II)Pd(II),\ Cu(II)Pd(II), Cu(II) are paramagnetic

Solution

Question 3

Cu2+Cu^{2+}Cu2+ ion has 3d93d^93d9 configuration while Zn2+Zn^{2+}Zn2+ ion has 3d103d^{10}3d10 configuration.

Cu2+Cu^{2+}Cu2+ ion has 3d53d^53d5 configuration while Zn2+Zn^{2+}Zn2+ ion has 3d63d^63d6 configuration.

Cu2+Cu^{2+}Cu2+ has half filled orbitals while Zn2+Zn^{2+}Zn2+ has fully filled orbitals.

CuSO4CuSO_4CuSO4​ is blue in colour while ZnSO4ZnSO_4ZnSO4​ is white.

Solution

Question 4

Cu2+>Ni2+>Cr3+>Fe3+Cu^{2+} > Ni^{2+} > Cr^{3+} > Fe^{3+}Cu2+>Ni2+>Cr3+>Fe3+

Ni2+>Cu2+>Fe3+>Cr3+Ni^{2+} > Cu^{2+} > Fe^{3+} > Cr^{3+}Ni2+>Cu2+>Fe3+>Cr3+

Fe3+>Cr3+>Ni2+>Cu2+Fe^{3+} > Cr^{3+} > Ni^{2+} > Cu^{2+}Fe3+>Cr3+>Ni2+>Cu2+

Cr3+>Fe3+>Ni2+>Cu2+Cr^{3+}> Fe^{3+} > Ni^{2+} > Cu^{2+}Cr3+>Fe3+>Ni2+>Cu2+

Solution

Question 5

1,41,41,4

3,43,43,4

2,32,32,3

1,21,21,2

Solution

Question 6

V2+V^{2+}V2+

Cr3+ Cr^{3+}Cr3+

Zn2+Zn^{2+}Zn2+

Ti3+Ti^{3+}Ti3+

Solution

Question 7

chromate obtained by the fusion of chromite ore with sodium carbonate in free access of air

pyrolusite which is fused with potassium hydroxide in the presence of air

iron pyrites by the fusion with potassium carbonate in presence of moisture

none of these

Solution

Question 8

K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​

(NH4)2[TiCI6](NH_4)_2[TiCI_6](NH4​)2​[TiCI6​]

VOSO4VOSO_4VOSO4​

K3[Cu(CN)4]K_3[Cu(CN)_4]K3​[Cu(CN)4​]

Solution

Question 9

adopt multiple oxidation states and to form complexes

form coloured ions

show paramagnetism due to unpaired electrons

form a large number of oxides

Solution

Question 10

(ii), (iii) and (iv)

(i), (iii) and (iv)

(i), (ii) and (iii)

(i), (ii) and (iv)

Solution

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$cu^{ 2+ }$

$Ni^{ 2+ }$

$Co^{2+}$

$Fe^{2+}$

$Sc(III),\ Ti(IV)$ are paramagnetic and $Pd(III),\ Cu(II)$ are diamagnetic

$Sc(III),\ Ti(IV)$ are diamagnetic and $Pd(II),\ Cu(II)$ are paramagnetic

$Cu^{2+}$ ion has $3d^9$ configuration while $Zn^{2+}$ ion has $3d^{10}$ configuration.

$Cu^{2+}$ ion has $3d^5$ configuration while $Zn^{2+}$ ion has $3d^6$ configuration.

$Cu^{2+}$ has half filled orbitals while $Zn^{2+}$ has fully filled orbitals.

$CuSO_4$ is blue in colour while $ZnSO_4$ is white.

$Cu^{2+} > Ni^{2+} > Cr^{3+} > Fe^{3+}$

$Ni^{2+} > Cu^{2+} > Fe^{3+} > Cr^{3+}$

$Fe^{3+} > Cr^{3+} > Ni^{2+} > Cu^{2+}$

$Cr^{3+}> Fe^{3+} > Ni^{2+} > Cu^{2+}$

$1,4$

$3,4$

$2,3$

$1,2$

$V^{2+}$

$Cr^{3+}$

$Zn^{2+}$

$Ti^{3+}$

$K_2Cr_2O_7$

$(NH_4)_2[TiCI_6]$

$VOSO_4$

$K_3[Cu(CN)_4]$